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AVprozaik [17]
2 years ago
13

A small hole in the wing of a space shuttle requires a 20.7-cm² patch.

Chemistry
1 answer:
cricket20 [7]2 years ago
8 0

The Patch's area of the space shuttle in km² is 2.07 × 10⁻⁹ km²

Given, that a space shuttle requires a 20.7 cm² patch

We have to convert the patch's area from cm² into km².

Unit conversion is a method in which we multiply or divide with a particular numerical factor and then finally round off to the nearest significant digits.

Patch area of the space shuttle is 20.7 cm²

1 cm = 0.00001 km

or, 1 cm² = (0.00001 km)²

or,  1 cm² = 10⁻¹⁰km²

20.7 cm² = 20.7 ×  10⁻¹⁰km²

20.7 cm² = 2.07 × 10⁻⁹ km²

The patch area in square kilometers is 2.07 × 10⁻⁹ km²

To learn more about unit conversion, visit: brainly.com/question/11543684

#SPJ4

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8 0
2 years ago
What element has the same amount of valence electrons as neon?<br> -KDM
Darya [45]

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Explanation:

3 0
3 years ago
Read 2 more answers
One liter of a buffer composed of 1.2 m hno2 and 0.8 m nano2 is mixed with 400 ml of 0.5 m naoh. what is the new ph? assume the
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Answer is: 3,4
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c₀(HNO₂) = 1,2 M = 1,2 mol/dm³.
c₀(NaNO₂) = 0,8 M = 0,8 mol/dm³.
V₀(HNO₂) = V₀(NaNO₂)  = 1 dm³ = 1 L.
c₀(NaOH) = 0,5 M = 0,5 mol/dm³.
n₀(HNO₂)= 1,2 mol/dm³ · 1 dm³ = 1,2 mol.
n₀(NaNO₂) = 0,8 mol/dm³ · 1 dm³ = 0,8 mol.
V(NaOH) = 400 mL · 0,001 dm³/mL = 0,4 dm³.
n₀(NaOH) = c₀(NaOH) · V₀(NaOH).
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n(HNO₂) = 1,2 mol - 0,2 mol = 1 mol.
n(NaNO₂) = 0,8 mol + 0,2 mol = 1 mol.
c(HNO₂) = 1 mol ÷ 1,4 dm³ = 0,714 mol/dm³.
c(NaNO₂) = 1 mol ÷ 1,4 dm³ = 0,714 mol/dm³.
pH = pKa + log (c(HNO₂) / c(NaNO₂)).
pH = 3,4 + log (0,714 mol/dm³ / 0,714 mol/dm³) = 3,4.
6 0
3 years ago
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